    # A case study for ζ(4)

Using symbolic summation tools in the setting of difference rings, we prove a two-parametric identity that relates rational approximations to ζ(4).

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## 1. Introduction

Kingdom: Mathematical constants
Class: Periods
Family: Multiple zeta values
Genus: Single zeta values
Species: Even zeta values

The quantity

 ζ(4)=∞∑k=11k4=π490

is a somewhat typical representative of even zeta values — the values of Riemann’s zeta function at positive even integers. It is shadowed by the far more famous , which was a main subject of Euler’s resolution of Basel’s problem, and  — an objet de l’étude of Apéry’s iconic proof of the irrationality of the latter (and also of ) [4, 30]. Though known to be irrational (and transcendental!), serves as a natural guinea pig for extending Apéry’s machinery to other zeta values. Apéry-type approximations to the number were discovered and rediscovered on several occasions [8, 29, 33], however they were not good enough to draw conclusions about its irrationality. An unexpected difficulty to control the ‘true’ arithmetic of those rational approximations to generated further research [14, 34], which eventually led to producing sufficient approximations and establishing a new world record for the irrationality measure of .

In this note we turn our attention to a rational side of the coin and prove the following two-parametric identity.

###### Theorem 1.

For integers , define two rational functions

 R(t)=Rn,m(t) =(−1)m(t+n2)(t−n)mm!(t−2n+m)2n−m(2n−m)! ×(t+n+1)n(t)n+1(t+n+1)2n−m(t)2n−m+1(n!(t)n+1)2 and ~R(t)=~Rn,m(t) =n!(t−n)2n−m(t)n+1(t)2n−m+1n∑j=0(nj)2(2n−m+jn)(t−j)nn!.

Then

 −13∞∑ν=n−m+1dR(t)dt∣∣∣t=ν=16∞∑ν=1d2~R(t)dt2∣∣∣t=ν. (1)

The instance of (1) was stated as Problem 1 in .

The fact that both sides of (1) are linear forms in and with rational coefficients is verifiable by standard techniques [14, 33, 34] which employ the partial-fraction decomposition of the rational functions. A remarkable outcome of this identity is the coincidence of two different-looking rational approximations to the zeta value. Such coincidences are often a source of deep algorithmic and analytical developments — check  for another exploration of this theme (see also ).

The main difficulty in establishing equality (1) (in contrast to tackling, for example, Apéry’s sums in  for ) is that its both sides are not hypergeometric functions but rather derivatives of hypergeometric functions. Another issue is that the summation range on the left-hand side is somewhat unnatural.

## 2. Symbolic summation

Denote by and the left- and right-hand sides of (1), respectively. In order to prove the identity (1) we proceed as follows.

(A) We compute the linear recurrence

 a0(n,m)Z(n,m)+a1(n,m)Z(n,m+1)+a2(n,m)Z(n,m+2)=0 (2)

with

 a0(n,m)=(2n−m)5,a1(n,m)=−(4n−2m−1)(6n4−24n3m+22n2m2−8nm3+m4−24n3+30n2m−14nm2+2m3+8n2−10nm+2m2−4n+m),a2(n,m)=−(2n−m−1)3(4n−m)(m+2), (3)

which holds simultaneously for and for all with .

(B) We show that the following initial values hold:

 Zl(n,0) =Zr(n,0)for alln≥0, (4) Zl(n,1) =Zr(n,1)for alln≥1. (5)

Combined with (A) this proves that holds true for all .

In order to carry out the steps (A) and (B), advanced symbolic summation techniques in the setting of difference rings are utilized. Among them the following three summation paradigms play a decisive role, that are available within the summation package Sigma .

(i) Creative telescoping. Given a sum and , one searches for constants , free of , and such that

 g(m,ν+1)−g(n,ν)=c0(n)f(m,ν)+c1(m)f(n+1,ν)+⋯+cδ(m)f(m+δ,ν) (6)

holds for all . Thus summing (6) over one obtains the recurrence

 g(m,b+1)−g(m,a)=c0(m)F(m)+c1(m)F(m+1)+⋯+cδ(m)F(m+δ). (7)

By specializing further — e.g., to and , or sending to if the limit exists — one obtains recurrence relations for more specific sums. The computed creative telescoping solution is also called a proof certificate for the recurrence (7) found: usually it allows one to verify that is a solution of (7) by simple polynomial arithmetic, without analyzing the usually complicated computation steps of the underlying summation algorithm. The algorithmic version of creative telescoping has been introduced in [32, 18] for hypergeometric sums. In order to prove (1), we will employ a generalized machinery for creative telescoping  where the summand can be composed not only in terms of hypergeometric products, but over indefinite nested sums defined over hypergeometric products. We emphasize that all recurrences produced below (using the Sigma-command GenerateRecurrence) are accompanied by such proof certificates which guarantee the correctness of all the calculations. Since the output is rather large and can be easily reproduced with Sigma, any explicit printout of the proof certificates is skipped.

(ii) Recurrence solving. Given a linear recurrence of the form (7), one can search for solutions that are expressible within certain classes function spaces. Using the Sigma-command SolveRecurrence one can search for hypergeometric solutions [17, 18] and, more generally, for all solutions that are expressible in terms of indefinite nested sums defined over hypergeometric products. Such solutions are also called d’Alembertian solutions [2, 19] a subclass of Liouvillian solutions .

(iii) Simplification of expressions. Within Sigma the expressions in terms of indefinite nested sums defined over hypergeometric products are represented in the setting of difference rings and fields [12, 23, 27]. Utilizing this difference ring machinery [24, 28] (compare also ) one can apply, e.g., the Sigma-command SigmaReduce to an expression in terms of indefinite nested sums. Then the output is a simplified expression where the arising sums and products (except products such as ) are independent among each other as functions of their external parameter. In particular, the input expression evaluates to zero (from a certain point on) if and only if Sigma reduces the expression to the zero-expression.

These summation paradigms can be used to transform a definite (multi-)sum to an expression in terms of indefinite nested sums by deriving a linear recurrence, solving the recurrence found in terms of indefinite nested sums, and, in case that sufficiently many solutions are found, combining them to an expression that evaluates to the same sequence as the input sum. Recently this machinery has been used for large scale problems coming from particle physics (see, e.g.,  and references therein). In this regard, also the package EvaluateMultiSum, which automatizes this summation mechanism, has been utilized non-trivially in the sections below.

## 3. A linear recurrence in m for the left-hand side

In order to activate the summation package Sigma, the sums arising in (1) have to be tailored to an appropriate input format. For this task, we split the left-hand side of (1) into the two subsums

 W1(n,m) =∞∑ν=2n−mdRn,m(t)dt∣∣∣t=ν=∞∑ν=0dRn,m(t+2n−m)dt∣∣∣t=ν and W2(n,m) =2n−m−1∑ν=n−m+1dRn,m(t)dt∣∣∣t=ν=n−1∑ν=1dRn,m(t+n−m)dt∣∣∣t=ν,

so that

 Zl(n,m)=−13(W1(n,m)+W2(n,m)). (8)

Observe that

 Rn,m(t+2n−m)=(−1)m(t+2n−m+n2)(t+n−m)mm!(t)2n−m(2n−m)!×(t+3n−m+1)n(t+2n−m)n+1(t+3n−m+1)2n−m(t)2n−m+1(n!(t+2n−m)n+1)2

and

 Rn,m(t+n−m)=(−1)m(t+n−m+n2)(t−m)mm!(t−n)2n−m(2n−m)!×(t+2n−m+1)n(t+n−m)n+1(t+2n−m+1)2n−m(t+n−m)2n−m+1(n!(t+n−m)n+1)2.

By definition all Pochhammer symbols in the former expression are of the form for some and . Thus, we can apply the well-known formula

 ddt(x+t)k∣∣t=ν=(x+ν)k(S1(ν+x+k−1)−S1(ν+x−1)) (9)

for with , where

 Sa(n)=n∑k=11ka

denotes the harmonic number of order . Employing this formula we get for all the following representation:

 F1(n,m,ν)=ddtRn,m(t+2n−m)∣∣∣t=ν =(−1)mn!2(1+ν)−1−m+2n(−m+n+ν)m(1−m+3n+ν)n(1−m+3n+ν)−m+2n2m!(−m+2n)!(−m+2n+ν)31+n(−m+2n+ν)1−m+2n ×(−6ν+ν(−2m+5n+2ν)(−S1(ν)−S1(−m+n+ν)+5S1(−m+2n+ν) −5S1(−m+3n+ν)−S1(−2m+4n+ν)+S1(n+ν)+S1(−m+4n+ν) +S1(−2m+5n+ν))+5n(m−2n)m−2n−ν+n(−2m+3n)n+ν+3n(m−n)−m+n+ν).

In addition, we will prepare the summand of . Here we note that the rule (9) cannot be applied to the arising factor , however we can easily overcome this issue by using the following: For with and any differentiable function , we have

 ddt((t−n)2n−mf(t))∣∣∣t=ν=(−1)n−νf(ν)(ν+n−m−1)!(n−ν)!.

Therefore, for all with we get

Because of the factor , we have for all with . In a nutshell, and can be written as

 W1(n,m)=∞∑ν=0F1(ν)andW2(n,m)=n−1∑ν=m+1F2(ν),

where the summands and are given in terms of hypergeometric products and linear combinations of harmonic numbers. Since these sums fit to the input class of Sigma, we can apply the command GenerateRecurrence to both sums and compute the recurrences

 a0(n,m)Ws(n,m)+a1(n,m)Ws(n,m+1)+a2(n,m)Ws(n,m+2)=rs(n,m)fors=1,2,

where the coefficients are given in (3) and where is too large to be reproduced here (verification of the latter equality required an extra simplification step with Sigma). To compute the recurrence for the hypergeometric sum one can alternatively use the Mathematica package fastZeil  based on .

Thus, given in (8) is a solution of the recurrence (2). For this part we needed 24 minutes to compute both recurrences and to combine them to (2).

## 4. A linear recurrence in m for the right-hand side

In order to calculate a linear recurrence for we will follow the same strategy as for in Section 3 by utilizing more advanced summation tools of Sigma. Define

 Gn,m(t)=n!(t−n)2n−m(t)n+1(t)2n−m+1(nj)2(2n−m+jn)(t−j)nn!

and write

 Zr(n,m) =16∞∑ν=1n∑j=0d2dt2Gn,m(t)∣∣∣t=ν =16(C1(n,m)+C2(n,m))

with

 C1(n,m)=∞∑ν=1n∑j=0d2dt2Gn,m(t+n)∣∣∣t=νandC2(n,m)=n∑ν=1n∑j=0d2dt2Gn,m(t)∣∣∣t=ν.

As before, we apply the formula (9) and its relatives to get the following monster summand of in terms of the harmonic numbers of order and :

 G0(n,m,j,ν)=d2dt2Gn,m(t+n)∣∣∣t=ν =(nj)2(j−m+2nn)(1+ν)−m+2n(1−j+ν+n)−1+n(1+ν+n)n(1+ν+n)−m+2n(ν+n)4(ν−m+2n)3 ×((ν+n)(ν−m+2n)(−ν(j−ν−n)(ν+n)(−1−j+ν+2n−S1(ν) +2S1(ν+n)−S1(ν+2n)−S1(ν−m+3n)−S1(−j+ν+n) +S1(ν−m+2n)+S1(−j+ν+2n)) −ν(j−ν−n)(ν−m+2n)(−1−j+ν+2n−S1(ν)+2S1(ν+n)−S1(ν+2n) −S1(ν−m+3n)−S1(−j+ν+n)+S1(ν−m+2n)+S1(−j+ν+2n)) +ν(ν+n)(ν−m+2n)(−1−j+ν+2n−S1(ν)+2S1(ν+n)−S1(ν+2n) −S1(ν−m+3n)−S1(−j+ν+n)+S1(ν−m+2n)+S1(−j+ν+2n)) −(j−ν−n)(ν+n)(ν−m+2n)(−1−j+ν+2n−S1(ν)+2S1(ν+n) −S1(ν+2n)−S1(ν−m+3n)−S1(−j+ν+n) +S1(ν−m+2n)+S1(−j+ν+2n)) +ν(j−ν−n)(ν+n)(ν−m+2n)(−1(j−ν−2n)2−S2(ν)+2S2(ν+n) −S2(ν+2n)−S2(ν−m+3n)−S2(−j+ν+n) +S2(ν−m+2n)+S2(−j+ν+2n)) +4(j+n)(ν+n)−3(ν+n)2+n(−m+n)−j(m+2n)) −2(ν+n)(−ν(j−ν−n)(ν+n)(ν−m+2n)(−1−j+ν+2n−S1(ν) +2S1(ν+n)−S1(ν+2n)−S1(ν−m+3n)−S1(−j+ν+n) +S1(ν−m+2n)+S1(−j+ν+2n)) −3(ν−m+2n)(−ν(j−ν−n)(ν+n)(ν−m+2n)(−1−j+ν+2n−S1(ν) +2S1(ν+n)−S1(ν+2n)−S1(ν−m+3n)−S1(−j+ν+n) +S1(ν−m+2n)+S1(−j+ν+2n)) +2jn(m−n)+2(j+n)(ν+n)2−(ν+n)3−(ν+n)(n(m−n)+j(m+2n))) −(ν+n)(ν−m+2n)(−ν(j−ν−n)(ν+n)(ν−m+2n)(−1−j+ν+2n −S1(ν)+2S1(ν+n)−S1(ν+2n)−S1(ν−m+3n)−S1(−j+ν+n) +S1(ν−m+2n)+S1(−j+ν+2n)) ×(−S1(ν+n)+S1(ν+2n)) +(ν+n)(ν−m+2n)(−ν(j−ν−n)(ν+n)(ν−m+2n)(−1−j+ν+2n −S1(ν)+2S1(ν+n)−S1(ν+2n)−S1(ν−m+3n)−S1(−j+ν+n) +S1(ν−m+2n)+S1(−j+ν+2n)) ×(−S1(ν)+S1(ν−m+2n)) −(ν+n)(ν−m+2n)(−ν(j−ν−n)(ν+n)(ν−m+2n)(−1−j+ν+2n −S1(ν)+2S1(ν+n)−S1(ν+2n)−S1(ν−m+3n)−S1(−j+ν+n) +S1(ν−m+2n)+S1(−j+ν+2n)) ×(−S1(ν+n)+S1(ν−m+3n)) +(ν+n)(ν−m+2n)(−ν(j−ν−n)(ν+n)(ν−m+2n)(−1−j+ν+2n −S1(ν)+2S1(ν+n)−S1(ν+2n)−S1(ν−m+3n)−S1(−j+ν+n) +S1(ν−m+2n)+S1(−j+ν+2n)) +2jn(m−n)+2(j+n)(ν+n)2−(ν+n)3 −(ν+n)(n(m−n)+j(m+2n))) ×(−1−j+ν+2n−S1(−j+ν+n)+S1(−j+ν+2n))).

In order to tackle the summand of , we have to differentiate twice. With and

 q(t)=Gn,m(t)p(t)=n!(t)n+1(t)2n−m+1(nj)2(2n−m+jn)(t−j)nn! (10)

we conclude that for all we have

 ~G(ν) =d2dt2Gn,m(t)∣∣∣t=ν=q(t)d2p(t)dt2+2dp(t)dtdq(t)dt+p(t)d2q(t)dt2∣∣∣t=ν =q(t)d2p(t)dt2+2dp(t)dtdq(t)dt∣∣∣t=ν;

the last equality follows since for all . In addition, we can use the following calculation: For and , we have

 d2dt2(t−n)2n−m∣∣∣t=ν=2ddt(Γ(t+n−m)(t−n)n−ν)∣∣∣t=ν.

In particular, if , we can apply the rule (9) to all Pochhammer symbols in (10) which gives

 G1(n,m,j,ν)=~G(ν) =2(−1)n+ν(nj)2(j−m+2nn)(1)n−ν(2)−1−m+n+ν(1−j+ν)−1+nν3(−m+n+ν)2(1+ν)n(1+ν)−m+2n ×(ν(−j+ν)(−m+n+ν)(1j−n−ν−S1(−j+ν)+S1(−j+n+ν)) +ν(−j+ν)(−m+n+ν)(−S1(−m+2n+ν)+S1(ν)) +ν(−j+ν)(−m+n+ν)(S1(ν)−S1(n+ν)) −ν(−j+ν)+ν(−m+n+ν) +2(j−ν)(−m+n+ν)+ν(−j+ν)(−m+n+ν) +ν(−j+ν)(−m+n+ν)(−1+S1(−m+n+ν)) −ν(−j+ν)(−m+n+ν)S1(n−ν)).

For , we use and the rule

 ddt((t−j)n)∣∣∣t=ν=f(ν)(−j+ν)−1+j−k+1(1−k+ν)−1−j+k+n

valid for any differentiable function , in place of (9). It follows that

 G2(n,m,j,ν)= ~G(ν)=(nj)2(2n−m+jn) ×2n!(ν+n−m−1)!(ν−n)n−ν(−j+ν)−1+j−k+1(1−k+ν)−1−j+k+nn!(t)n+1(t)2n−m+1.

Therefore,

 C2(n,m) =n∑ν=1n∑j=0d2dt2Gn,m(t)∣∣∣t=ν =n∑j=0n∑ν=max(j,1)G1(n,m,j,ν)+n∑j=0j−1∑ν=1G2(n,m,j,ν) =n∑j=1n∑ν=jG1(n,m,j,ν)+n∑ν=1G1(n,m,0,ν)+n∑j=0j−1∑ν=1G2(n,m,j,ν),

hence

 Zr(n,m) =16(C1(n,m)+C2(n,m)) =16(n∑j=0∞∑ν=1G0(n,m,j,ν)+n∑j=1n∑ν=jG1(n,m,j,ν) +n∑ν=1G1(n,m,0,ν)+n∑j=0j−1∑ν=1G2(n,m,j,ν)). (11)

Denote by , , , and the four resulting sums in (11) and use Sigma to compute four linear recurrences of with . A routine calculation demonstrates that each of the recurrences found can be brought to the form

 a0(n,m)As(n,m)+a1(n,m)As(n,m+1)+a2(n,m)As(n,m+2)=us(n,m), (12)

where the coefficients are given in (3) and where only the right-hand sides for differ. As an illustration, we provide with details about how we treat

 A0(n,m)=